What is the hybridization state of tin in tin(II) chloride, $\ce{SnCl2}$, and is the explanation similar for dichlorocarbene, $\ce{CCl2}$?

For $\ce{SnCl2}$, I read in my book that it is sp² but I am not able to explain it. I am not sure about the carbene.

I might have a misunderstanding about hybridisation, I think that the energy required for hybridization has to be compensated by the energy released in forming bond between hybridized orbitals.


2 Answers 2


Hybridisation is a mathematical concept, you never gain or lose energy due to it. Descriptions of the wave function involving hybrid orbitals or not are equally as valid. The linear combination of orbitals that result in the equilibrium geometry can however be interpreted as hybrid orbitals. Sometimes this may lead to a better understanding of the bonding situation.

From the above, it is obvious that hydbridisation is always a result from the molecular structure, never a cause for it. Hence we can deduce from the known bond angles the following:

Tin(II) chloride has a ∠Cl-Sn-Cl of about 95° (see Wikipedia). We know that p-orbitals are 90° towards each other. Therefore the tin will have more or less unhybrididised orbitals. See also this previous answer of me and this answer of Ben Norris.
Dichlorocarbene on the other hand has a ∠Cl-C-Cl of about 109° from both experimental and calculated data (Phys. Chem. Chem. Phys. 2002, 4 (14), 3282–3288). We know that the angles between sp³ hybrid orbitals are tetrahedral angles of about 109.5°. We can therefore deduce that the carbon employs sp³ hybrid orbitals in first order approximation.

The differences in the molecular structure and the utilisation of the s-orbitals for bonding stem from what is termed the inert pair effect, there is an explanation available in the this Q&A: What is the inert pair effect?

  • $\begingroup$ Re: "See also this answer of me and this answer of Ben Norris." Those answers talk about mainly the hydrides, while this question is about the chloride (of tin). Then, how can you say that the theory - that applies to hydride - will also apply to the chloride? Thanks for the answer! $\endgroup$ Feb 4, 2018 at 9:09
  • $\begingroup$ @GaurangTandon it applies in the way that the geometry is reason for a hybridisation concept and not the other way around. It doesn't matter which atoms are involved, it only matters how the atoms are arranged. (In this kind of approximation.) $\endgroup$ Feb 4, 2018 at 9:15
  • $\begingroup$ @Martin-マーチン Re: "the geometry is reason for a hybridisation concept" You mean to say that we deduce the hybridisation from the bond angle? If so, then, is it possible to predict the bond angle of $\ce{SnCl2}$ - i.e. whether it will be near $90^\circ$, $120^\circ$ or $180^\circ$ (only an approximation) - just from the atoms involved ($\ce{Sn}$ and $\ce{Cl}$)? $\endgroup$ Feb 4, 2018 at 9:32
  • $\begingroup$ @GaurangTandon the geometry depends on various factors, as such it is not possible to determine it from the atoms involved a priori. You'll have to use the tools at your disposal, be it experiment or quantum chemical analysis, even VSEPR may give you a good starting point. However, don't expect to much accuracy from the latter. $\endgroup$ Feb 4, 2018 at 9:48

1.The answer of question one is this that as Sn contain 4 electrons in its valence shell and will have to combine with two Cl atoms to form SnCl2 by sharing one electron with each Cl atom so it need two orbital which have one electron each then two free electrons will remain since there are maximum chances of these 2 free electrons to be present in the same orbital and be called as a lone pair hence Sn will undergo sp2 hybridization forming 3 sp2 hybridized orbitals and SnCl2 will have bent structure in which central atom Sn will be surrounded by 2 bond and 1 lone pair☺

  • 3
    $\begingroup$ I don't see why you post it after Martin's answer. $\endgroup$
    – Mithoron
    Jul 18, 2017 at 12:07
  • 1
    $\begingroup$ It's incorrect anyway. $\endgroup$
    – orthocresol
    Oct 1, 2018 at 12:31

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